F(y)=summation 1/(y^2+m^2) is not differentiable.

[R.P.F.]

Homework Statement

Is $$f(y)=\sum_{m=1}^\infty \frac{1}{y^2+m^2}$$differentiable?

Homework Equations

The Attempt at a Solution

From the graph, it is obvious that f is not differentiable at y=0, but I don't know how to prove that. I proved that $$\sum_{m=1}^n f_m=\sum_{m=1}^n\frac{1}{y^2+m^2}$$ converges uniformly to f. Does this help? Thank you very much.

 

[Dick]

The uniform convergence of differentiable functions does not prove the sum is differentiable. If the derivatives converge uniformly then it is. Don't you have a theorem like that? And why is it clear it's not differentiable at zero?

 

[R.P.F.]

The uniform convergence of differentiable functions does not prove the sum is differentiable. If the derivatives converge uniformly then it is. Don't you have a theorem like that? And why is it clear it's not differentiable at zero?

I have been thinking that each $$f_m$$ is strictly increasing on $$(0,\infty)$$. Therefore, so is f(y). If f is differentiable, then f'(0)=0 which is not true. Is this correct?

 

[Dick]

I have been thinking that each $$f_m$$ is strictly increasing on $$(0,\infty)$$. Therefore, so is f(y). If f is differentiable, then f'(0)=0 which is not true. Is this correct?

Do you mean decreasing? f'(0) has to be zero, the function is even. 1/(x^2+1) is strictly decreasing on [0,infinity). It's perfectly differentiable.

 

[R.P.F.]

Do you mean decreasing? f'(0) has to be zero, the function is even. 1/(x^2+1) is strictly decreasing on [0,infinity). It's perfectly differentiable.

Yeah I meant decreasing.
I know that each $$f_m$$ is differentiable on R. But f is not. I don't know how to prove this though. I am trying to prove by contradiction, but I cannot find a contradiction.

 

[R.P.F.]

Do you mean decreasing? f'(0) has to be zero, the function is even. 1/(x^2+1) is strictly decreasing on [0,infinity). It's perfectly differentiable.

I think that f is not differentiable at y=0 because the graph of f is really pointy at y=0...

 

[Dick]

I think that f is not differentiable at y=0 because the graph of f is really pointy at y=0...

If f(y) looks really pointy at y=0 it might be a problem with the graphing program. This is an analysis course, right? You shouldn't rely on that.

 

[R.P.F.]

If f(y) looks really pointy at y=0 it might be a problem with the graphing program. This is an analysis course, right? You shouldn't rely on that.

Hi,

Now I see how to show that it is differentiable at x=0, but still don't know how to show that it is differentiable on the rest of R. I have two tools: cauchy criterion and weierstrass m-test. I tried both but none worked. Can you give some hints? Thanks.

 

[Dick]

Hi,

Now I see how to show that it is differentiable at x=0, but still don't know how to show that it is differentiable on the rest of R. I have two tools: cauchy criterion and weierstrass m-test. I tried both but none worked. Can you give some hints? Thanks.

Like I said before, you want to prove the sum of the derivatives converges uniformly to f'(x). The weierstrass m-test should work fine. Can you show use where you are having trouble with it?

 

[R.P.F.]

Like I said before, you want to prove the sum of the derivatives converges uniformly to f'(x). The weierstrass m-test should work fine. Can you show use where you are having trouble with it?

Hey i actually figured it out! I did not think about using fundamental calc to find a bound at first...Thanks! :)